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3 years ago

Stars on the left of the diagram above are (View the picture to see the diagram)

Physics

2 answers:

omeli [17]3 years ago

4 0

<em>D. Less luminous than those on the right []</em>

oksian1 [2.3K]3 years ago

4 0

Answer:

D. Less luminous than those on the right []

Hope this helped

Explanation:

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The radii of curvature of a biconvex lens are 4 cm and 15 cm. The lens is in air, its index of refraction is 1.5. An object is a

rewona [7]

Answer:

v = 6.315 cm

Explanation:

given,

R₁ = 4 cm = 0.04 m

R₂ = 15 cm = 0.15 m

n =1.5

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

$\dfrac{1}{v}-\dfrac{1}{-1}=(1.5-1)(\dfrac{1}{0.04}+\dfrac{1}{0.15})$

$\dfrac{1}{v}+1 = 0.5 \times 31.66$

$\dfrac{1}{v} = 15.833$

v = 0.06315 m

v = 6.315 cm

hence, the distance of the image from the back surface is v = 6.315 cm

8 0

3 years ago

A GPS receiver can tell you your exact latitude and longitude?

Dmitry_Shevchenko [17]

That is false. A GPS cant tell latitude and longitude

3 0

3 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago

A 7 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The ki

Marizza181 [45]

Answer:

Explanation:

Given

mass of object $m=7 kg$

kinetic Energy $k=36\ J$

Tension in string $T=326\ N$

mass is moving in a horizontal circle so tension is providing the centripetal acceleration

therefore $T=\frac{mv^2}{r}----1$

where r=radius of circle

kinetic energy of particle $k=\frac{1}{2}mv^2----2$

divide 1 and 2 we get

$\frac{T}{k}=\frac{\frac{mv^2}{r}}{\frac{1}{2}mv^2}$

$\frac{T}{k}=\frac{2}{r}$

$r=\frac{2k}{T}=\frac{2\times 36}{326}$

$r=0.2208\ m$

8 0

3 years ago

Fast breeder reactor technology definition

mixas84 [53]

The definition is a nuclear reactor that generates more fissile material than it consumes.

5 0

3 years ago